9–40. Integration by parts Evaluate the following integrals us...

Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

40. ∫ e^√x dx

Accepted Answer

Below there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the following integral using integration by parts. The integral of sign of the square root of XDX. Awesome. So it appears for this particular problem we're asked to take the integral that is provided to us for the prom itself, and we're asked to evaluate it using integration by parts. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to we need to analyze this integrand using the acronym Lipit, which is L I P E T for short. So we're gonna be using lipid. Which, as you should recall, using lipid, the Integrand is Sign of the square root of X, and since this is a single function, it is often helpful to use a substitution method first. So let us say that T is equal to the square root of X, which is equal to x to the power of 1/2. Which will therefore simplify the integral. After substitution, we can therefore then recall and apply the integration by parts. So with that in mind, our second step is we need to perform the substitution. So we need to take T is equal to the square root of X, which we can say that X is equal to T the power of 2. Which we can therefore go on ahead to say that DX is equal to 2 TDT. Fantastic. So now, with that in mind, we now need to rewrite our integral. So we need to take the integral of Sign of the square root of X D X is equal to the integral of sine of T multiplied by 2 T D T is equal to 2 multiplied by the interval of T multiplied by sine of TDT, fantastic. So, moving right along here. Our third step is we need to recall and use Lipi again, L I P E T, our acronym Lippitt. To help us choose a value of U and DV. So once again we mentioned Lippit, but in case we're confused, let's define what Lipi stands for, so. L stands for logametic functions such as LN of X, which is the natural log of X. And I stands for, so once again, I stands for the inverse trigontric function, so such as inverse tangent of X. P stands for polynomial, which could be X to the power 2 or Y, etc. E stands for exponential and exponential such as E to the power of X. And finally, T stands for trigometric, such as sine of X or cosine of X, etc. So, using Lipi or acronym Lipi, we need to note that T stands for a polynomial, and sine of T stands for trigonometric. So according to our acronym Lipi, the polynomial P comes comes before the trigontric value. So that will mean that we will state that U is equal to T. So we could say that DU is equal to DT. And we need to choose DV is equal to sin of T DT, which we can therefore say that V is equal to negative cosine of T. Fantastic. So with that in mind, we now need to recall and apply the integration by parts, and that's our 4th step. So we could say that the integral of T multiplied by sine of TD T is equal to you. Multiplied by V minus the integral of VDU, which is equal to negative T multiplied by cosine of T, plus the integral of, so once again negative T multiplied by cosine a T plus the interval of cosine of T. DT, which is equal to negative T multiplied by cosine of T. Plus sign of T + C. Woohoo, we're on a roll. So now. Our next step, which is our 5th step, is we need to substitute our values back into the original equation and write our final answer, so we need to substitute everything back into its proper place. So we need to take the integral. Of sine of the square root of X D X is equal to 2 multiplied by negative T multiplied by cosine of T plus sign of T. Plus C, which will mean that our final answer will be all equal to, so our expression will be equal to -2 multiplied by the square root of X, multiplied by cosine of the square root of X. Plus 2 multiplied by sine of the square root of X plus C, and that will be our final answer. Hooray, we did it. So looking at our problem itself, to recap what we've learned, that will mean once again our final answer without a doubt will have to be the following, which once again is -2 multiplied by the square root of X, multiplied by. The cosine of the square root of X, plus 2 multiplied by the sign of the square root of X, plus C. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

9–40. Integration by parts Evaluate the following integrals using integration by parts.40. ∫ e^√x dx

Watch next

9–40. Integration by parts Evaluate the following integrals using integration by parts.
40. ∫ e^√x dx